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Prove that $\left(4\sqrt{2} + \frac{5}{3}\right)$ is an irrational number given that $\sqrt{2}$ is an irrational number.
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Let $4\sqrt{2} + \frac{5}{3}$ be a rational number.
$\therefore 4\sqrt{2} + \frac{5}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $$\begin{aligned}& b \neq 0 \\ & 4\sqrt{2} = \frac{a}{b} - \frac{5}{3} \\ & sqrt{2} = \frac{3a - 5b}{12b} \\ & 3a - 5b\end{aligned}$$ and $12b$ are integers.
$\therefore$ RHS is rational.
But LHS = $\sqrt{2}$ is an irrational number which is contradiction to our supposition.
Hence $4\sqrt{2} + \frac{5}{3}$ is an irrational number.
$\therefore 4\sqrt{2} + \frac{5}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $$\begin{aligned}& b \neq 0 \\ & 4\sqrt{2} = \frac{a}{b} - \frac{5}{3} \\ & sqrt{2} = \frac{3a - 5b}{12b} \\ & 3a - 5b\end{aligned}$$ and $12b$ are integers.
$\therefore$ RHS is rational.
But LHS = $\sqrt{2}$ is an irrational number which is contradiction to our supposition.
Hence $4\sqrt{2} + \frac{5}{3}$ is an irrational number.